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Question: A linearly polarized electromagnetic wave given as \[E = {E_0}\hat i\cos \left( {kz - \omega t} \rig...

A linearly polarized electromagnetic wave given as E=E0i^cos⁡(kz−ωt)E = {E_0}\hat i\cos \left( {kz - \omega t} \right) is incident normally on a perfectly reflecting infinite wall at z=az = a. Assuming that the material of the wall is optically inactive, the reflected wave will be given as
A. E⃗r=−E0i^cos⁡(kz−ωt){\vec E_r} = - {E_0}\hat i\cos \left( {kz - \omega t} \right)
B. E⃗r=E0i^cos⁡(kz+ωt){\vec E_r} = {E_0}\hat i\cos \left( {kz + \omega t} \right)
C. E⃗r=−E0i^sin⁡(kz+ωt){\vec E_r} = - {E_0}\hat i\sin \left( {kz + \omega t} \right)
D. E⃗r=E0i^sin⁡(kz−ωt){\vec E_r} = {E_0}\hat i\sin \left( {kz - \omega t} \right)

Explanation

Solution

The electromagnetic wave perfectly reflects when it is reflected from the denser medium. When the EM wave reflects from the denser medium, the type of the wave does not change but there will be change in phase by πc{\pi ^c}. After the reflection, the direction of propagation and direction of the electric field will get reversed.

Complete step by step answer:
As we have given that the wave is perfectly reflected. The electromagnetic wave perfectly reflects when it is reflected from the denser medium. We know that when the EM wave reflects from the denser medium, the type of the wave does not change, that is the transverse wave remains the transverse wave but there will be change in phase by πc{\pi ^c}.

Since the EM wave is incident normally on the reflecting wall, it will reflect back in the same direction with i^=−i^\hat i = - \hat i and z=−zz = - z creating the additional phase of πc{\pi ^c}.We have given the equation of the incident wave, E=E0i^cos⁡(kz−ωt)E = {E_0}\hat i\cos \left( {kz - \omega t} \right).
Therefore, we can write the reflected wave equation as,
E=E0(−i^)cos⁡(k(−z)−ωt+π)E = {E_0}\left( { - \hat i} \right)\cos \left( {k\left( { - z} \right) - \omega t + \pi } \right)
⇒E=−E0i^cos⁡(−(kz+ωt)+π)\Rightarrow E = - {E_0}\hat i\cos \left( { - \left( {kz + \omega t} \right) + \pi } \right)
⇒E=−E0i^cos⁡(π−(kz+ωt))\Rightarrow E = - {E_0}\hat i\cos \left( {\pi - \left( {kz + \omega t} \right)} \right)
We know the identity, cos⁡(π−θ)=−cos⁡θ\cos \left( {\pi - \theta } \right) = - \cos \theta .
Using this identity in the above equation, we get,
∴E=E0i^cos⁡(kz+ωt)\therefore E = {E_0}\hat i\cos \left( {kz + \omega t} \right)

So, the correct answer is option B.

Note: The term k in the given wave equation is the wavenumber of the EM wave and it is given as, k=2πλk = \dfrac{{2\pi }}{\lambda }, where, λ\lambda is the wavelength. In the given wave equation, the electric field is along the unit vector i^\hat i and the direction of propagation of the wave is along the z-axis. After the reflection, both the direction of the electric field and propagation change.